Question

Solve the initial-value problem. x' + 2tx = 5t, x(0) = 8 x(t) =

Answer 1

Multiply both sides of the ODE

`x'+2tx+5t`

by
`e^(t^2)`:

`e^(t^2)x'+2te^(t^2)x=5te^(t^2)`

Now the left side can be condensed as the derivative of a product:

`\left(e^(t^2)x\right)'=5te^(t^2)`

Integrate both sides, then solve for x :

`e^(t^2)x=\frac52e^(t^2)+C`

`\implies x(t)=\frac52+Ce^(-t^2)`

Given that x(0) = 8, we find

`8=\frac52+Ce^0\implies C=\frac{11}2`

so that the particular solution to this IVP is

`\boxed{x(t)=\frac{5+11e^(-t^2)}2}`